Question

Giải các hệ phương trình sau :

1, \(\left\{{}\begin{matrix}\left(y^2-8x+5\right)x^2+6y-5=0\\xy+1=x^2+y\end{matrix}\right.\)

2, \(\left\{{}\begin{matrix}x^2-xy+y^2=1\\x^2+2xy-y^2-3x-y=-2\end{matrix}\right.\)

( Tick cho mọi người nha ! )

Answer 1

a/ \(xy+1=x^2+y\Leftrightarrow x^2-1-xy+y=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)-y\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1-y\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\y=x+1\end{matrix}\right.\)

- Với \(x=1\Rightarrow y^2+6y-8=0\Rightarrow...\)

- Với \(y=x+1\)

\(\Rightarrow\left(x^2-6x+6\right)x^2+6x+1=0\)

\(\Leftrightarrow x^4-6x^3+6x^2+6x+1=0\)

Nhận thấy \(x=0\) ko phải nghiệm, chia 2 vế cho \(x^2\)

\(x^2+\frac{1}{x^2}-6\left(x-\frac{1}{x}\right)+6=0\)

Đặt \(x-\frac{1}{x}=t\Rightarrow x^2+\frac{1}{x^2}=t^2+2\)

\(\Rightarrow t^2+2-6t+6=0\Leftrightarrow t^2-6t+8=0\Leftrightarrow...\)

Answer 2

2/

Cộng vế với vế:

\(\Rightarrow2x^2+xy-3x-y+1=0\)

\(\Leftrightarrow\left(2x^2-3x+1\right)+y\left(x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)+y\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x-1+y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\2x+y=1\end{matrix}\right.\)

TH1: \(x=1\Rightarrow y^2-y=0\Rightarrow...\)

Th2: \(y=1-2x\)

\(\Rightarrow x^2-x\left(1-2x\right)+\left(1-2x\right)^2=1\)

\(\Leftrightarrow...\)